Scholarly article on topic 'Sharp Weak Type Inequality for Fractional Integral Operators Associated with d-Dimensional Walsh–Fourier Series'

Sharp Weak Type Inequality for Fractional Integral Operators Associated with d-Dimensional Walsh–Fourier Series Academic research paper on "Mathematics"

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Academic research paper on topic "Sharp Weak Type Inequality for Fractional Integral Operators Associated with d-Dimensional Walsh–Fourier Series"

Integr. Equ. Oper. Theory

DOI 10.1007/s00020-013-2116-7 |"T~ T"Z ~

© The Author(s) This article is published I Integral Equations

with open access at Springerlink.com 2013 I and Operator Theory

Sharp Weak Type Inequality for Fractional Integral Operators Associated with d-Dimensional Walsh-Fourier Series

Adam Osekowski

Abstract. Suppose that d > 1 is an integer, a E (0, d) is a fixed parameter and let Ia be the fractional integral operator associated with d-dimensional Walsh-Fourier series on [0,1)d. The paper contains the proof of the sharp weak-type estimate

2d — 1

IIIa (f)\\Ld/{d-a([0,1)d ) < (2d-a — 1)(2« — 1) IIf ([0,1)d ) •

The proof rests on Bellman-function-type method: the above estimate is deduced from the existence of a certain family of special functions.

Mathematics Subject Classification (2010). Primary 42B25, 42B30; Secondary 42B35.

Keywords. Fractional, best constant, weak-type inequality, Bellman function.

1. Introduction

Our motivation comes from the very natural question about sharp versions of estimates for d-dimensional Walsh system. As evidenced in numerous papers, such inequalities play an important role in many areas of mathematics, including approximation theory, Fourier analysis, harmonic analysis and probability theory. We refer the interested reader to the works [2,5,20,21,23,24] and references therein.

Let us start with introducing the necessary background and notation. We will work with functions defined on the unit cube [0,1)d in Rd, equipped with its dyadic sub-cubes, i.e., the sets of the form [fl, fin1) x [fj, a^) x '—[ fn, for some nonnegative integer n and some a\, a2, ..., ad E

{0,1,..., 2n — 1}. Recall that the Rademacher system {rn}n>o of functions on [0,1) is given by

rn(0=sgn (sin(2n+1 nt)).

IS? Birkhauser

Published online: 19 December 2013

Then {wn}n>o, the Walsh system on [0,1), is defined as follows: w0 = 1 and if n is a positive integer with n = 2ni + 2n2 + ... + 2nfc and n1 > n2 > ... > nk, then

Wn (t) = Tni (t)rn2 (t) ... Tnk (t) .

The d-dimensional counterpart of the Walsh system is the collection of all functions on [0,1)d which are of the form

x = (X1,X2, ...,Xd) ^ Wji (xi)Wj2 (x2 ) .. .Wjd (xd),

where j1, j2, ..., jd are nonnegative integers.

Now, assume that f is an integrable function on the cube [0,1)d. We define the associated rectangular partial sums of d-dimensional Walsh-Fourier series by the formula

ni —1 n2 — 1 nd-1 d

Sni,n2 ,...,nd (f)(x) = ... f(j1,j2,...,jd)Wjk(xk).

ji=0 j2 = 0 jd=0 k=1

Here x = (x1 ,x2,..., xd) G [0,1)d and

f (x) n Wjk (xk)dx

[0,1)d k = 1

is the (j1,j2,... ,jd)th Walsh-Fourier coefficient of f. The relation between the size of f and the behavior of the partial Sn¡n¡...¡n(f) has gained a lot of interest in the literature. For this and closely related topic, consult e.g. the works of Goginava [5,6], Goginava and Weisz [7], Nagy [9], Simon [13,14] and Weisz [21-23]. We will study this interplay from a slightly different point of view. Given a parameter a G (0, d), consider the associated fractional integral operator Ia by

Iaf = k"Sk,k,...,k(f).

This is the discrete and localized version of the usual fractional integral operator (Riesz potential) in

Rd (see Stein [17]). This object was studied by Watari [20] (a convenient reference, which presents a probabilistic approach, is the paper of Chao and Ombe [2]). For more recent works, we refer the interested reader to the works of Lacey et. al. [8] and Cruz-Uribe and Moen [3]. The arguments presented in these papers can be used to prove that the fractional integral operator is bounded as an operator from Lp([0,1)d) to Lq([0,1)d), where 1 < p < d/a and 1 = p — j. Furthermore, in the limit case p =1, q = d/(d — a), we have the weak-type estimate

\\Iaf llL9,~([0,1)d) < Ca,d\\f || LP ([0,1)d),

L„~([0,1)d) = supA\{x G [0,1)d : \f(x)\ > A}\1/«

is the usual weak-type quasi-norm. This should be compared to the analogous statements concerning the classical Riesz potentials on Rd; see e.g. Stein [17].

The principal purpose of the present paper is to derive the optimal value of the weak type constant Ca,d. Here is the precise statement.

Theorem 1.1. For any 0 < a < d and any f G Lp([0,1)d) we have

2d — 1

\\!af ||id/(d-o),^([0ji)d) < (2d-a — 1)(2a — 1) llf H^1 ([0,l)d) - (1.1) The inequality is sharp for any a and d.

The proof of the above statement will exploit an enhancement of the so-called Bellman function method. This technique originates from the theory of stochastic optimal control, and its connection with other areas of mathematics was firstly observed by Burkholder in [1], who studied certain sharp inequalities for martingale transforms. Since then, the method has been intensively developed in the subsequent works of Burkholder and his students (a convenient reference on the subject is the monograph [12] by the author). Furthermore, in the late 1990s, Nazarov, Treil and Volberg [10,11] showed that the method can be exploited in a much wider analytic context. Since the seminal papers, the technique has been used in numerous settings: see e.g. [4,15,16,18,19] and references therein.

Roughly speaking, the Bellman function method enables to deduce a given inequality from the existence of a certain special function, which enjoys some majorization and convexity-type properties. It turns out that in the study of (1.1), one needs an appropriate extension of the method. More precisely, the weak-type inequality will be deduced from the existence of a certain family of special functions. As we hope, this novel modification can be used in the investigation of other related estimates which arise naturally in the area.

A few words about the organization of the paper are in order. The special functions are introduced and studied in the next section. Then, in Section 3, we show how to exploit their properties to obtain the inequality (1.1). The final part of the paper contains the construction of certain extremal functions on [0,1)d, which show that for each a and d, the constant Ca}d cannot be improved.

2. A Special Function and its Properties

As announced in the previous section, the proof of (1.1) will depend heavily on the existence of certain special functions. Given a nonnegative integer n, consider Bn : {(x,y) : y > 2-nax > 0}^ [0,1] given by the formula

Bn(x, y)

(2d - 1)x2"

-(n+1)a

(2d - 1)x2-(n+1)a + (1 - y)(2d-a - 1)_

d/(d-a)

if y < 1, and Bn(x,y) = 1 otherwise.

The key convexity-type property of the family {Bn}n>0 is described in the following statement.

Then for any numbers h\, h2, .hd G [—x, (2d — 1)x] satisfying 5^k=1 hk = 0

Lemma 2.1. Let n > 1 be a fixed integer and fix x, y satisfying 0 < 2 uax < y.

en for we have

B (x y) >

Bu-i(x ,y) > ^Y. Bu(x + hk, y + 2-uahk). (2.1)

If y > 1, then the above estimate is obvious, since the left-hand side is equal to 1, while all the summands appearing on the left are at most 1. Therefore, we may assume that y < 1. We consider two cases.

Proof of (2.1) for y + 2-ua(2d — 1)x < 1. Then we have y + hk < 1 for all k. Consider the function

£(h) := Bu(x + h,y + 2-uah), h G [—x, (2d — 1)x].

Let us show that the graph of £ lies below the line segment joining (— x, £(—x)) and ((2d — 1)x, £((2d — 1)x)). Since the point

(0, Bu-i(x,y)) = (1 — 2-d) • (—x,£(—x)) + 2-d • ((2d — 1)x,£((2d — 1)x))

lies on the segment, this will immediately yield (2.1).

To prove the above statement, it suffices to prove the following:

(a) The left-sided derivative ((2d — 1)x) is larger or equal to the slope of the segment.

(b) We have £"(h) > 0 for h close to —x.

(c) The derivative changes its sign at most once.

We start with the property (a). Since £(— x) = 0, the slope of the segment joining (—x,£(—x)) and ((2d — 1)x, £((2d — 1)x)) equals

£((2d — 1)x)

1 [ (2d — 1)2d-(u+1)ax Td/(d-a)

= 2x |_(2d — 1)2d-(u+1)ax +(1 — y — 2-ua(2d — 1)x)(2d-a —

On the other hand, some tedious, but straightforward calculations show that the left-sided derivative of £ at (2d — 1)x is equal to

(2d — 1)2d-(u+1)ax 1a/(d-a)

d — a L(2d — 1)2d-(u+1)ax + (1 — y — 2-ua(2d — 1)x)(2d-a — 1)_

(2d — 1)2-(u+1)a(2d-a — 1)(1 — y + 2-uax) [(2d — 1)2d-(u+1)ax + (1 — y — 2-ua(2d — 1)x)(2d-a — 1)]2 '

Therefore, the assertion of (a) is equivalent to

d (2d-a — 1)(1 — y + 2-uax)

d — a (2d — 1)2d-(u+1)ax + (1 — y — 2-ua(2d — 1)x)(2d-a — 1): or, after some manipulations,

(d — a)(2a — 1)2d-(u+1)ax < a(1 — y + 2-uax)(2d-a — 1).

However, we have 1 — y > 2-na(2d — 1)x (this is the assumption under which we work: see the beginning of the proof). Therefore, we will be done if we show that

(d - a)(2a - 1)2d-(n+1)ax < a2

(2d-a - 1),

or, equivalently,

1 _ 2-a 2d-a — 2-a - < -.

This follows directly from the estimate d > a.

We turn our attention to (b) and (c). First write

£(h) =

f 2d - 1 ^ \2a - 1)

d/(d-a)

2-na(1 - 2-a)x - M' M + 2-na(1 - 2-a)h_

d/(d-a)

C [1 + g(h)]d/(d-a),

where M = (2d - 1)x2"

-(n+1)a

+ (1 - y)(2d-a - 1). Therefore,

(1 + g(h))(2

a-d)/(d-a)

However, we derive that

(g'(h))2 + (1+g(h))g"(h)

2-na(1 - 2-a)(M - 2-na(1 - 2-a)x) (M + 2-na(1 - 2-a)h)2

■) —2nc

g'(h) = -2

(1 - 2-a)^M - 2-na(1 - 2-a)x) (M + 2-na(1 - 2-a)h)3

Consequently, the sign of £''(h) is that of

-(M - 2-na(1 - 2-a)x - 2 • 2-na(1 - 2-a)(x + h).

Now both (b) and (c) follow at once, since M > 2 na (1 — 2 a)x and the above expression is a decreasing linear function of h.

Proof of (2.1) for y + 2-na(2d - 1)x > 1. As previously, put

That is,

£(h) =

C(h) := Bn(x + h,y + 2-nah), h G [-x, (2d - 1)x].

(2d - 1)(x + h)2-(n+1)a (2d - 1)(x + h)2-(n+1)a + (1 - y - 2-nah)(2d-a - 1)

d/(d-a)

if h < (1 — y)2na, and £(h) = 1 otherwise. The argument is a slight modification of that used above. First, let us show that the graph of £ lies below the line passing through the points (—x,£(—x)) = (—x, 0) and ((1 — y)2na, £((1 — y)2na)) = ((1 — y)2na, 1). The slope of the line is positive, so the majorization is obvious on [(1 — y)2na, (2d — 1)x]. Thus, it suffices to focus on the interval [—x, (1 — y)2na], for which it is enough to show that

(a) The left-sided derivative ((1 — y)2na) is larger or equal to the slope of the line.

(b) We have £"(h) > 0 for h close to —x.

(c) The derivative changes its sign at most once on (— x, (1 — y)2na).

Actually, the conditions (b) and (c) are proved by repeating the calculations from in the previous case, word-by-word (the formula for £(h) for h e [—x, (1 — y)2na] is the same). To show (a), we derive that the slope of the line is (x + (1 — y)2na) —1, while the left-sided derivative is given by

d (2d — 2a)

d — a (2d — 1)(x +(1— y)2na)' Therefore, the assertion of (a) can be rewritten as

2d — 2a 2d — 1

which follows directly from the estimate 0 < a < d. This implies the majorization of the graph of £. To complete the proof, we will show that the point (0,Bn-1(x,y)) lies above the line joining (— x,£(—x)) and ((1 — y)2na,£((1 — y)2na). This amounts to saying that Bn-1(x,y) is not smaller than

(1 — y)2na :Z(—x) + ^((1 — y)2na) = x

x +(1 — y)2na ' x +(1 — y)2no- ' ' x + (1— y)2na'

Now, the substitution w := (1 — y)2na/x turns the desired inequality into

2d—a _ 1 \ d/(d-«)

1+w> 1+w

2^ — 1

However, the right-hand side is a convex function of w, and both sides are equal if w = 0 or w = 2d — 1. It remains to note that 0 < (1— y)2na/x < 2d — 1 (the left estimate is trivial, the right follows from the assumption under which we work: see the beginning of the proof). □

We conclude this section by the following simple statement.

Lemma 2.2. (i) If n is a nonnegative integer and (x,y) lies in the domain of Bn, then Bn(x,y) > X{y>i}-(ii) We have

f 2d - 1 \ d/(d—a) Bo(x,x) < ^J xd/(d—a). (2.2)

Proof. (i) This follows directly from the definition. (ii) We have

(Bo(x x))(d—a)/d = _(2d — 1)2—"x_ < 2d — 1 x

(B0(X, x)) 2d—a — 1+ x(1 — 2 —«) < 2d — 2« ^

which is the claim. □

3. Proof of (1.1)

For any n > 0, let Fn denote the a-algebra generated by all dyadic subcubes of [0,1)d which are of measure 2-nd. It is convenient to split the reasoning into several separate parts.

Step 1. Some reductions. By a straightforward approximation argument, it is enough to prove the weak-type estimate for functions which are simple. Here by simplicity we mean that f is measurable with respect to some Fn (that is, the corresponding sequence (Sn,n,... ,n(f))n>o stabilizes after a finite number of steps). Introduce the difference operators (Dn)n>o given by Do(f) = So,o,...,o(f) and

Dn(f) = Sn,n,...,n(f) — Sn-1,n-1,...,n-1(f^ n = 11, 2 ....

Note that for any n > 0 and any atom A of Fn we have the equality

J Dn+i(f )(x)dx = 0. (3.1)

To see this, note that by the very definition, Dn+1f (x) is a linear combination of the products d=1 Wjk (xk) such that at least one of jk's is equal to n + 1. The corresponding Walsh function Wjk, integrated over a dyadic interval of length 2-n, gives 0 and hence (3.1) follows. In the probabilistic language, (3.1) means that (Sn n. n(f ))n>o, considered as a sequence of random variables (on the probability space ([0,1)d, B([0,1)d), |-|)), is a martingale with respect to the dyadic filtration (Fn)n>o. As an immediate consequence, we see that it suffices to show (1.1) for nonnegative functions only. Indeed, the passage from f to | f | does not affect the L1-norm; on the other hand, for any n > 0 we have

ISn,n,-,n(f)l = |E(f |Fn)| < E(|f||Fn) = Sn,n,...,n(If |),

so |Ia(f)| < Iadf|) and hence HIa(f)||L,,~([o,1)d> < Uladf|)||L?.~([o,1)d). Thus, from now on, we assume that f > 0.

Step 2. An alternative definition of Ia(f). It will be convenient to express the integral fractional operator in terms of the difference sequence. Namely, note that

I a (f ) = Y.2-naSn,n,-,n (f )

oo n <x

= ^Y.2-na Dk(f) = (1 — 2-a)-1 2-kaDk (f),

n=o k=o k=o

after the change of the order of summation. In what follows, we will use the notation

fn = Sn,n,... n(f) = Y, Dk (f), gn = YJ2-kaDk (f), n > 0. (3.2)

k=o k=o

Note that both (fn)n>0, (gn)n>0 are simple: for sufficiently large n we have fn = f and gn = (1 — 2-a)Ia(f). We conclude this part by proving that for any n > 0 we have

gn > 2-nafn. (3.3)

This follows from an easy induction. Indeed, for n = 0 we have g0 = f0; now, assuming that the bound holds true for n, we derive that

gn+1 = gn + dgn+l > 2-nafn + 2-(n+1)adfn+i > 2-(n+1)afn+i.

Step 3. The use of functions Bn, n >0. We will prove the following statement: for n 0,

J Bn(fn(s),gn(s))ds < J Bn+i(fn+i(s),gn+i(s))ds. (3.4)

[0,1)d [0,1)d

Note that the above formula makes sense: by (3.3), the pair (fn,gn) takes values in the domain of Bn.

To show (3.4), fix an atom A of Fn. Both fn and gn are Fn-measurable, and hence constant on A: denote the corresponding values by x and y, respectively. On the other hand, let A1, A2, ..., A2d be the pairwise disjoint atoms of Fn+1 contained in A. Then Dn+1(f) is constant on each Aj; put hj = Dn+1(f )\Aj. We have x + hj = fn\a, > 0, so

hj > —x for all j. (3.5)

Furthermore, (3.1) implies that

Y,hk =0, (3.6)

which, combined with (3.5), yields

hj = hk < (2d - 1)x. (3.7)

Thus the assumptions of Lemma 2.1 are satisfied and (2.1) gives

Bn(x,y) >2dY. Bn+i(x + hk ,y + 2-nahk), k=1

which is equivalent to

J Bn(fn(s),gn(s))ds> J Bn+i(fn+i(s),gn+i(s))ds. A A

It remains to sum over all A to obtain (3.4).

Step 4. Proof of (1.1). Recall that by simplicity of g, there is n such that gn = (1 — 2-a)Ia(f). Now, all that is left is to use Lemma 2.2. Using the first

part, then (3.4) and finally the second part of that lemma, we obtain

j X{(l-2-a)Ia (f )>!}ds = j X|g„(s)>l}ds [0,1)d [0,1)d

< J Bn(fn(s),gn(s))ds

[0,1)d

< J Bo(fo(s),go(s))ds

[o,1)d

/ id 1 \ d/(d-a)

< 2 - M II f |d/(d-a)

< I 2d - 2^ f "l1 ([0 'i)d)'

Here in the last passage we have exploited the fact that f0 is identically \\f ||Li([0 i)d) on [0,1)d. Now apply the above estimate to the function (1 — 2-a)-1 f/A and multiply both sides by Ad/(d-a). Taking the supremum over A on the left completes the proof of (1.1).

Remark 3.1. An important observation is in order. Namely, it is possible to base the above reasoning on one special function B only, instead of the whole family (Bn)n>0. Indeed, as one easily verifies, we have the equality

Bn(x,y) = B(2-nax, y), (3.8)

where B : {(x, y) : y > x > 0}^ [0,1] is given by

B{x,y) =

(2d-a - 2-a)x

d/(d-a)

(2d-a — 2-a)x +(1 — y)(2d-a — 1)_

if y < 1, and B(x,y) = 1 otherwise. Then the main inequality (2.1) takes the following more natural form in terms of B:

B(2ax, y) > 2dYl B(x + hk, y + hk),

if 0 < x < y, 0 < x + hk < y + hk and ^k=i = 0. The reason why we have decided to present the more complicated approach using the whole family (B„)„>o is that the above argumentation can be used for general problems, in which the special functions do not enjoy any scaling property of the type (3.8). Furthermore, the technicalities arising in both settings are of comparable complexity and the reduction (3.8) does not really simplify the calculations.

4. Sharpness

Now we will construct appropriate extremal example. Let d > 1, N > 1 be fixed integers and consider the function f = X[0 , 2-N)x[o, 2-N)x...x[o, 2-N) on [0,1)d. Furthermore, for any 0 < n < N, put An = [0, 2-n) x [0, 2-n) x _ x

[0, 2 n). Then for each such n, An is the unique atom of Fn containing AN, and therefore

S f (x) = \An 1 = 2(n-N )d

Sn,n,...,nJ \x) I A I ^ \An\

provided x G An, and Sn,n...,nf(x) = 0 elsewhere; furthermore, we have Sn,n,...,nf = Sn,n,...,nf = f for n > N. These facts can be deduced directly from the very definition of Snn..,n, or, alternatively, follow from the martingale interpretation of the sequence (Sn,n,...,nf )n>0, mentioned in the previous section. Using the above formulas for Sn,n,...,nf, we see that if x G AN, then

Ia(f )(x) = Y, 2-naSn,n,..,nf (x) = E 2-na+(n-N)d + ^ 2-na

n=0 n=0 n=N+1

Nd 2(N +1)(d-a) _ 1 2"(N+1)a

2d-a _ 1 1 _ 2-a

Consequently,

NT- ,«„ f2-Nd 2(N+ 1)(d-a)-1 + 2-(N+1)a \ I A \(d-a)/d

\\Ia (I )\\.Ld/(d-a),~([o,1)d) ' 2d a 1 + 1-2-a ) \AN \

\\f\\L1([0,1)d) \AN \

2-N(d-a)(2(N+1)(d-a) _ 1) 2-a

2d-a _ 1 1_2-a' However, the integer N was arbitrary; letting N ^<x>, we see that the latter expression converges to

2d-a 2-a 2d — 1 +

2d-a _ 1 1 _ 2-a (2d-a _ 1)(2a _ 1)' This shows that the constant in (1.1) is indeed the best possible.

Acknowledgements

The author would like to thank the referee for the careful reading of the paper and helpful suggestions, which in particular led to Remark 3.1 above. The results were obtained when the author was visiting Purdue University, USA. The research was supported in part by the NCN grant DEC-2012/05/B/ST1/00412.

Open Access. This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.

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[24] Zygmund, A.: Trigonometric Series, vol. 2. Cambride University Press, London (1968)

Adam Osekowski (b) Department of Mathematics Informatics and Mechanics University of Warsaw Banacha 2 02-097 Warsaw Poland

e-mail: ados@mimuw.edu.pl

Received: September 29, 2013. Revised: December 1, 2013.